Expectation of Continuous Uniform Distribution
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Theorem
Let $a, b \in \R$ such that $a < b$.
Let $X \sim \ContinuousUniform a b$ be the continuous uniform distribution over $\closedint a b$.
Then:
- $\expect X = \dfrac {a + b} 2$
Proof
From the definition of the continuous uniform distribution, $X$ has probability density function:
- $\map {f_X} x = \begin {cases} \dfrac 1 {b - a} & : a \le x \le b \\ 0 & : \text {otherwise} \end {cases}$
From the definition of the expected value of a continuous random variable:
- $\ds \expect X = \int_{-\infty}^\infty x \map {f_X} x \rd x$
So:
\(\ds \expect X\) | \(=\) | \(\ds \int_{-\infty}^a 0 x \rd x + \int_a^b \frac x {b - a} \rd x + \int_b^\infty 0 x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \intlimits {\frac {x^2} {2 \paren {b - a} } } a b\) | Primitive of Power, Fundamental Theorem of Calculus | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {b^2 - a^2} {2 \paren {b - a} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {b - a} \paren {b + a} } {2 \paren {b - a} }\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a + b} 2\) |
$\blacksquare$
Also see
Sources
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Appendix $13$: Probability distributions
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $15$: Probability distributions